# If x = 4t1+t2, y = 3(1-t21+t2), then show that dydx=-9x4y - Mathematics and Statistics

Sum

If x = (4"t")/(1 + "t"^2), y = 3((1 - "t"^2)/(1 + "t"^2)), then show that ("d"y)/("d"x) = (-9x)/(4y)

#### Solution

x = (4"t")/(1 + "t"^2)

Differentiating both sides w.r.t. ‘t’, we get

("d"x)/"dt" = "d"/"dt" ((4"t")/(1 + "t"^2))

= ((1 + "t"^2)*"d"/"dt"(4"t") - 4"t"*"d"/"dt"(1 + "t"^2))/(1 + "t"^2)^2

= ((1 + "t"^2)(4) - 4"t"(0 + 2"t"))/(1 + "t"^2)^2

= (4 + 4"t"^2 - 8"t"^2)/(1 + "t"^2)^2

= (4 - 4"t"^2)/(1 + "t"^2)^2

= (4(1 - "t"^2))/(1 + "t"^2)^2

y = 3((1 - "t"^2)/(1 + "t"^2))

("d"y)/"dt" = 3*"d"/"dt"((1 - "t"^2)/(1 + "t"^2))

= 3[((1 + "t"^2)*"d"/"dt"(1 - "t"^2) - (1 - "t"^2)*"d"/"dt"(1 + "t"^2))/(1 + "t"^2)]

= 3[((1 + "t"^2)(0 - 2"t") - (1 - "t"^2)(0 + 2"t"))/(1 + "t"^2)^2]

= 3[(-2"t"(1 + "t"^2) - 2"t"(1 - "t"^2))/(1 + "t"^2)^2]

= 3(- 2"t")[(1 + "t"^2 + 1 - "t"^2)/(1 + "t"^2)^2]

= - 6"t" xx 2/(1 + "t"^2)^2

= (-12"t")/(1 + "t"^2)^2

∴ ("d"y)/("d"x) = (("d"y)/("dt"))/(("d"x)/("dt"))

= ((-12"t")/((1 + "t"^2)^2))/((4(1 - "t"^2))/((1 + "t"^2)^2)

= ("d"y)/("d"x) = (-3"t")/(1 - "t"^2)    ......(i)

Also, (-9x)/(4y) = (-9((4"t")/(1 + "t"^2)))/(4 xx 3((1 - "t"^2)/(1 + "t"^2))

= (-3"t")/(1 -"t"^2)     ......(ii)

From (i) and (ii), we get

("d"y)/("d"x) = (-9x)/(4y)

Concept: Derivatives of Parametric Functions
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Chapter 1.3: Differentiation - Q.5
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